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Tidal forces on the Sun [extracted from Sun and Barycentre Period]

The exact conjunction of all the planets is very rare and the same is true of groups of conjunctions so the roughly 178 year pattern will not exactly repeat, I don't know if it ever does and the spiral diagram of the relative positions of the sun centre and barycentre illustrates the 2D complexity, let alone the 3D complexity. I have been interested in the tidal effects these movements might and indeed must cause inside the sun. The recent pictures of Jupiter and its vortices shows how the moons stir Jupiter, plus the internal heat, so Jupiter is like an analog of the sun situation.

Last edited by grapes; 2018-Feb-08 at 06:04 PM.
Reason: Reference url

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

If we are interested in the possibility that the planets might have a physical role in the Sun's activity, let's avoid falling into the trap of thinking the barycentric displacement of the Sun is the key quantity. Any physical stress on the Sun is in the form of tidal stretch, and for any given planetary mass this falls off as the cube of the distance to the planet, while the barycentric loop increases with distance. A remote planet puts greatly reduced gravitational action on the Sun but keeps it up a very long time, thus causing the Sun to move in a relatively large loop.

These are for mean distances. Note that Venus exceeds Jupiter's average amount, and that Saturn contributes less than the amount that Mercury varies in its highly eccentric orbit. Let me add that any two planets will raise the same amount of spring tide when on opposite sides of the sun as when in conjunction on the same side, regardless of where the overall barycenter is. For this problem, assuming the tidal action has any effect on the circulation of the Sun's material, my inclination is to concentrate on the first three inner planets and Jupiter and ignore everything else. It is my opinion that the barycentric excursion is merely a mathematical curiosity when analyzing action in and very near the Sun. For gravitational action on Voyager 2 and outlying comets and KBOs, by all means take it into account.

If we are interested in the possibility that the planets might have a physical role in the Sun's activity, let's avoid falling into the trap of thinking the barycentric displacement of the Sun is the key quantity. Any physical stress on the Sun is in the form of tidal stretch, and for any given planetary mass this falls off as the cube of the distance to the planet, while the barycentric loop increases with distance. A remote planet puts greatly reduced gravitational action on the Sun but keeps it up a very long time, thus causing the Sun to move in a relatively large loop.

These are for mean distances. Note that Venus exceeds Jupiter's average amount, and that Saturn contributes less than the amount that Mercury varies in its highly eccentric orbit. Let me add that any two planets will raise the same amount of spring tide when on opposite sides of the sun as when in conjunction on the same side, regardless of where the overall barycenter is. For this problem, assuming the tidal action has any effect on the circulation of the Sun's material, my inclination is to concentrate on the first three inner planets and Jupiter and ignore everything else. It is my opinion that the barycentric excursion is merely a mathematical curiosity when analyzing action in and very near the Sun. For gravitational action on Voyager 2 and outlying comets and KBOs, by all means take it into account.

Actually, because the sun is fluid, a point mass on the sun is being attracted toward say Jupiter by Newtonian gravity inverse distance squared as the sun spins, not uniformly as a solid would, but at different rates. This is like the oceans on earth pulled sideways by the moon (and sun) however the Suns radial heat flows are greatly complicated and dominated by magnetic forces on the plasma,. My idea was that the variation of the tidal effect as the sun spins disrupted radial flow but annoyingly the Jupiter effect swing s from in phase to out of phase with the sunspot cycle. So I don't know whether that forcing has an effect as a damped oscillation.

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

Actually, because the sun is fluid, a point mass on the sun is being attracted toward say Jupiter by Newtonian gravity inverse distance squared as the sun spins, not uniformly as a solid would, but at different rates. This is like the oceans on earth pulled sideways by the moon (and sun) however the Suns radial heat flows are greatly complicated and dominated by magnetic forces on the plasma,. My idea was that the variation of the tidal effect as the sun spins disrupted radial flow but annoyingly the Jupiter effect swing s from in phase to out of phase with the sunspot cycle. So I don't know whether that forcing has an effect as a damped oscillation.

My bold. That is what tidal stretching is all about, that is, the difference between the gravitational "pull" on the near side and the far side. It drops off as the cube of the distance, not the square, which explains why the giant planets seem so weak compared to the inner terrestrials. The lack of correlation between the position of Jupiter and the sunspot activity suggests that the effect of the tidal action is too slight to affect the circulation of the Sun's innards with their tangles of magnetic fields.

Brace yourselves for just how feeble the tidal action of the planets on the Sun really is. I estimated that the Earth's component will elongate the Sun by about a trillionth of its diameter, or roughly a millimeter in over a million kilometers. Of course I did not do a rigorous equipontential surface calculation because I don't have that kind of knowhow, but I did a sanity check with the same calculation for the Earth and Moon, and my result was about the same order of magnitude as the tides we actually observe. With all of the planets and the Sun in syzygy, giving the ultimate compound spring tide, we get a few millimeters. I would say the physical effect on the Sun in comparison with its Coriolis effect and magnetic entanglements is utterly negligible.

Yes that is right for the normal radial tidal movement which is tiny. But I propose that you miss the tangential point for fluids just like the ocean tides which are sideways accelerations at constant radius. The gravity pull integrates up for approximately the quater turn of the sun, then negligible for another quarter turn , then it reverses. I am not talking about the tiny radial lift which is negligible compared with the gravity of the main body, sun or earth. It's a tiny pull maintained for a long time and not resisted in a fluid while, of course it is resisted by a solid. Fluids accelerate and pick up velocity and Coriolis applies too.

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

Numbers, taking G as 6.67 10^-11, Jupiter mass and distance 1.9.10^27 kg 778.10^6 km,
Using Newton
The force on one kg is 2.10^-3 N
Acceleration same in m/s^2

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

Click on the link, and then on the PDF tab that appears in lower left corner.

The author's rigorous calculation of an equipotential displacement in a fluid Earth-sized body gave the same order of magnitude as my oversimplified rough-and-dirty estimate. Thus I stand by my estimate for the Sun and my expectation that physical consequences of the tides raised by the planets will be vanishingly small.

Oh dear, I hate to disagree but the tidal calculation using the near and far side is not what I am talking about . that radial tidal bulge along the two body centre line is indeed tiny.
I hope I can explain again.
If we consider a particle at the surface, free to move at constant radius because it is floating, it sees a sideways gravitation pull toward, let's say just one distant body for moment, Newton gives us the equation just as that paper does but we do not need to consider r.
I choose to consider a particle at the extreme outside, ninety degrees from the centre line between the bodies. So this small force is at right angles to the main gravity force holding the particle to the body.
That particles sees an inverse square sideways force and it is free to move sideways toward the distant body.
The earth case is easier and it was Euler who spotted this important way to analyse the tides.
The major tidal force is the effect of those tangential accelerations as the earth rotates.
The sun case is obviously more complex because a gas is even freer than a water ocean and it churns so I accept the tides may still be insignificant, but not so insignificant as that cube of the distance suggests.

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

Maclaurin used Newton's theory to show that a smooth sphere covered by a sufficiently deep ocean under the tidal force of a single deforming body is a prolate spheroid (essentially a three-dimensional oval) with major axis directed toward the deforming body. Maclaurin was the first to write about the Earth's rotational effects on motion. Euler realized that the tidal force's horizontal component (more than the vertical) drives the tide. In 1744 Jean le Rond d'Alembert studied tidal equations for the atmosphere which did not include rotation.

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

Maclaurin used Newton's theory to show that a smooth sphere covered by a sufficiently deep ocean under the tidal force of a single deforming body is a prolate spheroid (essentially a three-dimensional oval) with major axis directed toward the deforming body. Maclaurin was the first to write about the Earth's rotational effects on motion. Euler realized that the tidal force's horizontal component (more than the vertical) drives the tide. In 1744 Jean le Rond d'Alembert studied tidal equations for the atmosphere which did not include rotation.

i extracted that from the WP article and was in haste, forgot to attribute it. There are other references to Euler who first pointed out that the separate calculation of the horizontal forces is important for fluid tides but would not be necessary for any solid. The sun is of course rotating at various rates and that is the important factor.

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

Oh dear, I hate to disagree but the tidal calculation using the near and far side is not what I am talking about . that radial tidal bulge along the two body centre line is indeed tiny.
I hope I can explain again.
If we consider a particle at the surface, free to move at constant radius because it is floating, it sees a sideways gravitation pull toward, let's say just one distant body for moment, Newton gives us the equation just as that paper does but we do not need to consider r.
I choose to consider a particle at the extreme outside, ninety degrees from the centre line between the bodies. So this small force is at right angles to the main gravity force holding the particle to the body.
That particles sees an inverse square sideways force and it is free to move sideways toward the distant body.
The earth case is easier and it was Euler who spotted this important way to analyse the tides.
The major tidal force is the effect of those tangential accelerations as the earth rotates.
The sun case is obviously more complex because a gas is even freer than a water ocean and it churns so I accept the tides may still be insignificant, but not so insignificant as that cube of the distance suggests.

My bold. That makes no sense to me. In the author's derivation leading up to equation 20, the inverse cube of the distance to the perturbing body clearly appears in the calculations for the strength of the tidal action. In that section of the paper it was my understanding that he was envisioning fluid bodies, which when rotating come closer to the equipotential surface than do rigid ones. Therefore whatever applies to our oceans seems to be a reasonable analogy to what is happening in the Sun.

I just now ran equation 20, substituting the Sun for the Earth, and the Earth for the Moon. I came up with an equipotential elongation of about 45mm in the direction of the Earth and a compression of half that at quadrature. I can see I previously underestimated the action by perhaps a couple of orders of magnitude. It serves me right for trying to do it in my head while doing an aerobic power walk on a treadmill. It is still proportionately tiny compared with the Moon's action on the Earth. We have an equipotential elongation of 1/8 the amount on the Earth, spread out over a body 109 times the radius. With all of the planets in syzygy, we would get 6 or 7 times as much. Venus would be the biggest contributor with Jupiter a close second on the average.

Suppose for the sake of argument we propose that this tidal action has a significant effect on the Sun's convection with its entangled magnetic fields. I would look for the synodic period of Jupiter with respect to Venus. The spring tide/neap tide pattern would have large ripples from Mercury and Earth, small ones from Mars and Saturn, and vanishingly small ones from Uranus and Neptune.

My view now is that the tides are not significant despite the closeness of the sunspot cycle to Jupiter orbit it's not close enough. However there must be tides. Euler added to the bulge argument. The horizontal movement at the tangent cannot be denied, it is a basic application of gravity to a point mass. I am not at all sure that the barycentre should
Be used on the calculation the Newtonian formula applies to the bodies not the mass centre.
In terms of effect Venus is as powerful as Jupiter but faster of course.

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

My view now is that the tides are not significant despite the closeness of the sunspot cycle to Jupiter orbit it's not close enough. However there must be tides. Euler added to the bulge argument. The horizontal movement at the tangent cannot be denied, it is a basic application of gravity to a point mass. I am not at all sure that the barycentre should
Be used on the calculation the Newtonian formula applies to the bodies not the mass centre.
In terms of effect Venus is as powerful as Jupiter but faster of course.

My bold. I am not denying it. I think it is my understanding that it explains why the ocean excursion in places far from land comes closer to the equipotential surface than does the underlying solid body.

We have a spring tide/neap tide cycle on the Sun with a period several months from Jupiter and Venus, and its amplitude will vary with Jupiter's position in its eccentric orbit. As mentioned before, Mercury and Earth add their own sizable components, complicating the issue.

It appears that Jose fell into a trap of envisioning the large barycentric displacements of the Sun by the outermost giant planets as a significant physical disturber of the Sun's internal state. As mentioned before, these motions are of large amplitude but are very slow, and even if they were faster the Sun would not "feel" them anymore than an astronaut in free fall feels Earth's gravity. It only feels the tidal action, which will feel the same whether it is from a small nearby planet that causes small but quick barycentric cycles, or from a large and distant planet that causes larger but slower cycles. What it feels is only a function of its rotation and the magnitude of the tidal action.

Yes that is right for the normal radial tidal movement which is tiny. But I propose that you miss the tangential point for fluids just like the ocean tides which are sideways accelerations at constant radius. The gravity pull integrates up for approximately the quater turn of the sun, then negligible for another quarter turn , then it reverses. I am not talking about the tiny radial lift which is negligible compared with the gravity of the main body, sun or earth. It's a tiny pull maintained for a long time and not resisted in a fluid while, of course it is resisted by a solid. Fluids accelerate and pick up velocity and Coriolis applies too.

No that's not true at all.

Originally Posted by profloater

Oh dear, I hate to disagree but the tidal calculation using the near and far side is not what I am talking about . that radial tidal bulge along the two body centre line is indeed tiny.
I hope I can explain again.
If we consider a particle at the surface, free to move at constant radius because it is floating, it sees a sideways gravitation pull toward, let's say just one distant body for moment, Newton gives us the equation just as that paper does but we do not need to consider r.
I choose to consider a particle at the extreme outside, ninety degrees from the centre line between the bodies. So this small force is at right angles to the main gravity force holding the particle to the body.
That particles sees an inverse square sideways force and it is free to move sideways toward the distant body.

No, not relative to the body that it is on.

The earth case is easier and it was Euler who spotted this important way to analyse the tides.
The major tidal force is the effect of those tangential accelerations as the earth rotates.
The sun case is obviously more complex because a gas is even freer than a water ocean and it churns so I accept the tides may still be insignificant, but not so insignificant as that cube of the distance suggests.

Perhaps important is that there is not a tangential acceleration at the tangent point itself. One must understand that all tidal forces on the Sun must be registered in the frame of the Sun, which is already a frame accelerating toward the distant source of gravity in question. That overall acceleration of the entire Sun produces a kind of "fictitious force" in the Sun's frame that exactly cancels the tangential acceleration at the tangent point, leaving only the radial acceleration there. The direction that the tidal forces point can be seen at https://www.lockhaven.edu/~dsimanek/scenario/stress.gif . However, this fact is only an "order unity" correction to profloater's point. Hornblower is right that the deformation of the equipotential is very small, yet there's also the issue of how far the particles need to move to get to the equipotential. Still, this is a gas, not a liquid, so the particles don't need to move nearly as far as they would have to for a liquid of constant density. I'm on the side of inconsequential effects.

I agree it is tiny as well as complex but I stand with Euler who was the first to point out the tangential effect in relation to earth ocean tides. Using a fixed sun reference the sun sees Jupiter orbiting it as the biggest influence. Mentioning the barycentre thread I still maintain that if there are sun behaviours cyclic with the planets they will be mass over distance squared gravity effects plus tinier mass over distance cubed torque or tidal effects and not anything to do with the mass balance calculation of barycentre. A particle on the surface feels such a huge radial gravity force from the sun centre that a tangential pull from Venus or Jupiter seems inconsequential and that's without going into the huge magnetic forces.

sicut vis videre estoWhen we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.Originally Posted by Ken G

Hornblower found that Jupiter was indeed the dominant effect, and notice that the tangential acceleration you are talking about is still an effect that scales with the inverse distance cubed. I'm sure Euler got that also. If you think it is an inverse distance squared effect, you are still forgetting to subtract the acceleration at the center of the Sun due to Jupiter, or whatever planet is being considered. It is necessary to subtract that, because the tidal affect appears in the frame of reference of the Sun, which is an accelerated reference frame, so must have the fictitious forces subtracted to get the correct tidal effects-- even for tangential forces. See the figure at https://www.lockhaven.edu/~dsimanek/scenario/stress.gif for the direction these tangential forces point, and bear in mind they are inverse cube forces.

To repeat a point I made earlier, the Sun does not "feel" the overall acceleration toward a planet any more than an astronaut in low Earth orbit feels the gravitational acceleration which is nearly as strong as at sea level. It is only the distortion of the equipotential surface that it feels, which is proportional to the difference in the near side and far side gravitational acceleration. That difference is inverse cube dependent.